A quadrilateral PQRS has an inscribed circle, the points of tangencies with sides PQ, QR, RS, SP being A, B, C, D, respectively. Let the midpoints of AB, BC, CD, DA be E, F, G, H, respectively. Prove that the angle between segments PR and QS is equal to the angle between segments EG and FH.
Problem
Source: Taiwan 2nd TST 2005, 3rd independent study, problem 2
Tags: geometry, geometry solved
21.08.2005 15:33
k2c901_1 wrote: A quadrilateral PQRS has an inscribed circle, the points of tangencies with sides PQ, QR, RS, SP being A,B,C,D, respectively. Let the midpoints of AB, BC, CD, DA be E,F,G,H, respectively. Prove that the angle between segments PR and QS is equal to the angle between segments EG and FH. We will work with directed angles modulo 180°. Let I be the center of the incircle of the quadrilateral PQRS. Since the incircle touches the sides SP and PQ of the quadrilateral PQRS at the points D and A, these points D and A are symmetric to each other with respect to the line PI. Thus, DA⊥PI, and the midpoint H of the segment DA lies on the line PI. Similarly, AB⊥QI, BC⊥RI, CD⊥SI, and the midpoints E, F, G of the segments AB, BC, CD lie on the lines QI, RI, SI, respectively. Since DA⊥PI, we have < IHD = 90°; equivalently, < IHD = - 90° (since 90° = - 90° as we are using directed angles modulo 180°). On the other hand, since the incircle of the quadrilateral PQRS, centered at I, touches the side SP at the point D, we have ID⊥SP and thus < IDP = 90°. Hence, < IHD = - < IDP. Also, trivially, < HID = - < DIP. Thus, the triangles IHD and IDP are inversely similar. Therefore, IHID=IDIP, or, equivalently, IH⋅IP=ID2. Similarly, IF⋅IR=IB2. But ID = IB, since the points D and B both lie on the incircle of the quadrilateral PQRS, which is centered at I; thus, we get IH⋅IP=IF⋅IR. Hence, by the converse of the intersecting chords theorem, it follows that the points H, P, F and R lie on one circle. Thus, < PRF = < PHF; equivalently, < (PR; RI) = < (PI; FH). Similarly, < (QS; SI) = < (QI; EG). On the other hand, since BC⊥RI and CD⊥SI, we have < (RI; BC) = 90° and < (SI; CD) = 90°, so that < (RI; SI) = < (RI; BC) + < (BC; CD) - < (SI; CD) = 90° + < (BC; CD) - 90° = < (BC; CD) = < BCD. But similarly, < (QI; PI) = < BAD, and since the points A, B, C, D lie on one circle (namely, on the incircle of the quadrilateral PQRS), we have < BCD = < BAD. Thus, < (RI; SI) = < (QI; PI). Hence, < (PR; QS) = < (PR; RI) + < (RI; SI) - < (QS; SI) = < (PI; FH) + < (QI; PI) - < (QI; EG) = < (PI; FH) + < (EG; PI) = < (EG; FH). And the problem is solved. darij